In a triangle $ABC, AD$ is the altitude from $A\text{ (Fig. 12.11). Given }b>c,\mathrm{\angle }C={23}^{\circ }$and $AD=\frac{abc}{b^2-c^2}$ then $\mathrm{\angle }B\mathit{.}$ is equal to

We have $AD=b\sin C$

$\begin{array}{r}\Rightarrow \frac{abc}{b^2-c^2}=b\sin C\\ \Rightarrow \frac{\sin C}{c}=\frac{a}{b^2-c^2}\\ \Rightarrow \frac{\sin A}{a}=\frac{a}{b^2-c^2}\end{array}$

$\Rightarrow \sin A=\frac{a^2}{b^2{c}^2}\left[\because \frac{\sin C}{c}=\frac{\sin A}{a}\right]$

           $\begin{array}{l}=\frac{{\sin }^{2}A}{{\sin }^{2}B-{\sin }^{2}C}\\ =\frac{{\sin }^{2}A}{\sin (B+C)\sin (B-C)}\end{array}$

$$\begin{gathered} \begin{array}{l}\Rightarrow \sin (B-C)=1 [\because \sin (B+C)=\sin A]\\ \Rightarrow B-C={90}^{\circ }\\ \end{array} \\ B={90}^{\circ }+C={90}^{\circ }+{23}^{\circ }={113}^{\circ } \end{gathered}$$